Question

A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by dollar 3. find the original duration of the tour. (Hint use quadratic equation)

Answer 1

Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for $x$ days.

So, for $x$ days the total expense = $360

Thus the daily expense in dollars can be given by = $\frac{360}{x}$

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

Tour duration now = $(x+4)$ days

On extension, his daily expense is cut by $3

New daily expense in dollars = $(\frac{360}{x}-3)$

Total expense in dollars can now be given as:  $(x+4)(\frac{360}{x}-3)$

Simplifying by using distribution (FOIL).

$(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))$

$360-3x+\frac{1440}{x}-12$

$348-3x+\frac{1440}{x}$

We know total expense remains the same which is = $360.

So, we have the equation as:

$348-3x+\frac{1440}{x}=360$

Multiplying each term with $x$ to remove fractions.

$348x-3x^2+1440=360x$

Subtracting $348x$ both sides

$348x-348x-3x^2+1440=360x-348x$

$-3x^2+1440=12x$

Dividing each term with -3.

$\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}$

$x^2-480=-4x$

Adding $4x$ both sides.

$x^2+4x-480=-4x+4x$

$x^2+4x-480=0$

Solving using quadratic formula.

For a quadratic equation: $ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plugging in values from the equation we got.

$x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}$

$x=\frac{-4\pm\sqrt{16+1920}}{2}$

$x=\frac{-4\pm\sqrt{1936}}{2}$

$x=\frac{-4\pm44}{2}$

So, we have

$x=\frac{-4+44}{2}$   and $x=\frac{-4-44}{2}$

$x=\frac{40}{2}$   and $x=\frac{-48}{2}$

∴ $x=20$           and $x=-24$

Since number of days cannot be negative, so we take $x=20$ as the solution for the equation.

Thus, the original duration of the tour = 20 days